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QJEEM22Ch4ade: JEE Main 2022 Chemistry - Class 12 JEEM22Ch4ade Solution: If [Cu(H 2 O) 4 ] 2+ absorbs a light of wavelength 600 nm for…

If [Cu(H2O)4]2+ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+ will be ____________ ×\times 10 −- 21 J. [Nearest Integer]

(Given : h = 6.63 ×\times 10 −- 34 Js and c = 3.08 ×\times 108 ms −- 1)

Step-by-Step Verified Solution

[Cu(H2O)4]2+\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+} is tetrahedral

[Cu(H2O)6]2+\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} is octahedral

∵Δt=49×Δ0\because \Delta_{\mathrm{t}}=\frac{4}{9} \times \Delta_{0}

amp;Δt=6.63×10−34×3.08×108600×10−9amp;Δ0=94×6.63×10−34×3.08×108600×10−9≈765.7×10−21 J\begin{aligned}&\Delta_{\mathrm{t}}=\frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \\\\&\Delta_{0}=\frac{9}{4} \times \frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \approx 765.7 \times 10^{-21} \mathrm{~J}\end{aligned}

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